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ajprincess

  • 2 years ago

Plzzz help. the following figure shows a long wire carrying a current of 30A. The rectangular loop carries a current of 20A. Calculate the resultant force acting on the loop. Assume that a=1cm, b=8cm and l=30cm.

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  1. ajprincess
    • 2 years ago
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    |dw:1337319857835:dw|

  2. timo86m
    • 2 years ago
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    What is B?

  3. ajprincess
    • 2 years ago
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    u mean my answer for B?

  4. timo86m
    • 2 years ago
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    I think it will involve integration.

  5. ajprincess
    • 2 years ago
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    |dw:1337320632924:dw| This is to find the force on the second current conductor.

  6. ajprincess
    • 2 years ago
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    Sorry so bad at drawing. :(

  7. timo86m
    • 2 years ago
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    ah ok i think i found it in book i wil lattach 2 files hold on

  8. timo86m
    • 2 years ago
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    1 Attachment
  9. timo86m
    • 2 years ago
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    1 Attachment
  10. timo86m
    • 2 years ago
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    hope it helps

  11. ajprincess
    • 2 years ago
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    Ya. it helps a lot. Bt y hav they given the value of l. It is askd to find the resultant force.

  12. timo86m
    • 2 years ago
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    i is amps. Is that what you mean?

  13. ajprincess
    • 2 years ago
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    No. In ur attachment the question is to fin the magnetic force on the top of the wire b. And in the equation given in the solution there is no l bt only a nd b r there. in my question the value of l is also given. I am wondering y that was given and the magnetic force askd to b found here is the resultant.

  14. ajprincess
    • 2 years ago
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    i mean top of the wire of length b.

  15. timo86m
    • 2 years ago
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    Not sure just solve for whatever variable or convert. :) but the diagram is almost exactly the same :)

  16. ajprincess
    • 2 years ago
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    ha k. thannnnxxxxxxxx a lot.

  17. timo86m
    • 2 years ago
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    :)

  18. ujjwal
    • 2 years ago
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    Is your answer equal to 0.0032 N ?

  19. ajprincess
    • 2 years ago
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    i get a different answer. Bt can u pls show me hw u got ur answer if u dnt mind?

  20. ajprincess
    • 2 years ago
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    @ujjwal

  21. ujjwal
    • 2 years ago
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    sure, the |dw:1337350437168:dw| there exists an attractive force F1 and a repulsive force F2 between the wires carrying 30 amp and 20 amp.. Calculate both forces and find their difference.. that's all..

  22. ajprincess
    • 2 years ago
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    which formula did u use here to find it?

  23. ajprincess
    • 2 years ago
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    did u round it off?

  24. ajprincess
    • 2 years ago
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    @ujjwal

  25. ajprincess
    • 2 years ago
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    i got the answer as 0.00315N.

  26. ujjwal
    • 2 years ago
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    It involves large numbers.. your answer seems to be correct.. I used the relation\[F=mu _{0}I _{1} I _{2}l/2pi r \]

  27. ujjwal
    • 2 years ago
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    what's wrong with OS? but i guess you got the relation i used...

  28. ajprincess
    • 2 years ago
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    ya. do u knw how to derive this relation from this |dw:1337397048799:dw|

  29. ajprincess
    • 2 years ago
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    |dw:1337397220582:dw|

  30. ujjwal
    • 2 years ago
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    I guess i can't see the whole relation.. I can't see it at rightmost part.. Is there a lower limit?? It's a line integral in closed path , isn't it? why do you need limits?? I can't see the whole relation though..

  31. ajprincess
    • 2 years ago
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    |dw:1337397679038:dw|

  32. ajprincess
    • 2 years ago
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    |dw:1337397784654:dw| This formula is to find the force on the second conductor. Can u @Shivam_bhalla explain hw it becomes |dw:1337397890531:dw|. Plzzzz.

  33. shivam_bhalla
    • 2 years ago
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    @ajprincess , Take a look here -->http://cnx.org/content/m31103/latest/

  34. shivam_bhalla
    • 2 years ago
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    The above link for magnetic field due to a wire and its derivation. Now I will explain that concept which you want me to do so.

  35. shivam_bhalla
    • 2 years ago
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    |dw:1337353084406:dw| Consider 2 parallel wires carrying current I_1 and I_2 Force exerted by wire 1 on wire 2 is F= (B_1) (I_2)(distance between two wires)

  36. shivam_bhalla
    • 2 years ago
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    Sorry I wrote that in a hurry. It should be F= (B_1) (I_2)(length of wire 2) Now as in the previous link I gave to you, we see that \[B_1 = \frac { \mu_0 I_1 }{2 \pi r}\] \[F=B_1I_2(length \space of \space wire \space 2)\] \[\frac{F}{length \space of \space wire \space 2} = \frac { \mu_0 I_1 }{2 \pi r} I_2\] \[f=\frac { \mu_0 I_1I_2}{2 \pi r} \] where\[ f = \frac{F}{length \space of \space wire \space 2}\]

  37. ajprincess
    • 2 years ago
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    Thanxxxx a lot @Shivam_bhalla

  38. ajprincess
    • 2 years ago
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    Thanxx a lot @ujjwal

  39. shivam_bhalla
    • 2 years ago
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    yw :)

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